Download, print and study this document offline 
Page 1 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s =  64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \  = +  64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t   = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = +  ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = 60m 6.1s = 9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration =  + Final velocity initial velocity s ( ) 4 8 =  0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under vt graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20  Mechanics1 h +20 m/s – 60 m v time 12s 8s 4s v max q Page 2 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s =  64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \  = +  64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t   = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = +  ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = 60m 6.1s = 9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration =  + Final velocity initial velocity s ( ) 4 8 =  0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under vt graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20  Mechanics1 h +20 m/s – 60 m v time 12s 8s 4s v max q (b)As the particle did not return back distance travelled in 12 s = Displacement at 12 s \ Average speed = 8 m/s. 6. (a) Radius ( ) R of circle = 21 22 m \ Circumference of circle = 2pR = ´ ´ 2 22 7 21 22 m =6 m Speed ( ) v of particle = 1 m/s \ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved = ´ = = ° 2 6 2 2 3 120 p p Magnitude of Average velocity = = Magnitude of displacement Time s ( ) 2 = = ° AB R 2 2 60 2 sin =R 3 2 = 21 22 3 2 = 21 3 44 m/s (b) Magnitude of average acceleration =  ½ ½ ½ ½ ½ ½ ½ ½ ® ® v v B A 2s = ° 2 120 2 2 vsin [Q     v v A B v ® ® = = = 1 m/s) = ° vsin 60 = 3 2 m/s 2 7. Position vector at t = 0 s r i j 1 1 2 ® = + ( ) ^ ^ m Position vector at t = 4 s r i j 2 6 4 ® = + ( ) ^ ^ m (a) Displacement from t = 0 s to t = 4 s =  ® ® ( ) r r 2 1 = +  + ( ) ( ) ^ ^ ^ ^ 6 4 1 2 i j i j = + ( ) ^ ^ 5 2 i j m \ Average velocity = + ( ) ^ ^ 5 2 4 i j m s = + ( . . ) ^ ^ 125 05 i j m/s (b) Average acceleration =  Final velocity Initial velocity s 4 = +  + ( ) ( ) ^ ^ ^ ^ 2 10 4 6 4 i j i j =  + 2 4 4 i j ^ ^ =  + ( . ) ^ ^ 05i j m/s 2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement between A and B is 10 m 1 2 1 2 1 10 2 2 gt gt   = ( ) or t t 2 2 1 2   = ( ) or t t t 2 2 2 1 2   + = ( ) or 2 3 t = t =1.5 s 9. The two bodies will meet if Displacement of first after attaining = highest point Displacement of second before attaining highest point Motion in One Dimension  21 120° O B A(t = 0 s) (t = 2s) v B ® v A ® Page 3 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s =  64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \  = +  64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t   = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = +  ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = 60m 6.1s = 9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration =  + Final velocity initial velocity s ( ) 4 8 =  0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under vt graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20  Mechanics1 h +20 m/s – 60 m v time 12s 8s 4s v max q (b)As the particle did not return back distance travelled in 12 s = Displacement at 12 s \ Average speed = 8 m/s. 6. (a) Radius ( ) R of circle = 21 22 m \ Circumference of circle = 2pR = ´ ´ 2 22 7 21 22 m =6 m Speed ( ) v of particle = 1 m/s \ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved = ´ = = ° 2 6 2 2 3 120 p p Magnitude of Average velocity = = Magnitude of displacement Time s ( ) 2 = = ° AB R 2 2 60 2 sin =R 3 2 = 21 22 3 2 = 21 3 44 m/s (b) Magnitude of average acceleration =  ½ ½ ½ ½ ½ ½ ½ ½ ® ® v v B A 2s = ° 2 120 2 2 vsin [Q     v v A B v ® ® = = = 1 m/s) = ° vsin 60 = 3 2 m/s 2 7. Position vector at t = 0 s r i j 1 1 2 ® = + ( ) ^ ^ m Position vector at t = 4 s r i j 2 6 4 ® = + ( ) ^ ^ m (a) Displacement from t = 0 s to t = 4 s =  ® ® ( ) r r 2 1 = +  + ( ) ( ) ^ ^ ^ ^ 6 4 1 2 i j i j = + ( ) ^ ^ 5 2 i j m \ Average velocity = + ( ) ^ ^ 5 2 4 i j m s = + ( . . ) ^ ^ 125 05 i j m/s (b) Average acceleration =  Final velocity Initial velocity s 4 = +  + ( ) ( ) ^ ^ ^ ^ 2 10 4 6 4 i j i j =  + 2 4 4 i j ^ ^ =  + ( . ) ^ ^ 05i j m/s 2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement between A and B is 10 m 1 2 1 2 1 10 2 2 gt gt   = ( ) or t t 2 2 1 2   = ( ) or t t t 2 2 2 1 2   + = ( ) or 2 3 t = t =1.5 s 9. The two bodies will meet if Displacement of first after attaining = highest point Displacement of second before attaining highest point Motion in One Dimension  21 120° O B A(t = 0 s) (t = 2s) v B ® v A ® v t gt v t t g t t 0 2 0 0 0 2 1 2 1 2  =    ( ) ( ) or 0 1 2 2 0 0 0 2 0 =    v t g t t t ( ) or gt t gt v t 0 0 2 0 0 1 2 = + or t gt v g = + 1 2 0 0 = + t v g 0 0 2 10. 5 1 2 2 = gt Þ t =1s For A H t gt = + 0 1 2 2 . or H gt = 1 2 2 …(i) For B H g t  =  25 1 2 1 2 ( ) …(ii) or 1 2 1 2 1 25 2 2 gt gt   = ( ) [Substituting value of H from Eq. (i)] 1 2 1 25 2 2 g t t [ ( ) ]   = t t t 2 2 2 1 5   + = ( ) 2 1 5 t  = Þ t = 3 s Substituting t = 3 s in Eq. (i) H = ´ ´ = 1 2 10 3 45 2 m 11. s at = 1 2 0 2 Forward motion v at = 0 Backward mo tion  = +  s at t a t ( ) ( ) 0 2 1 2 \  =  1 2 1 2 0 2 0 2 at at t at ( ) or  =  t t t t 0 2 0 2 2 or t t t t 2 0 0 2 2 0   = t t t t = ±    2 2 41 2 0 0 2 0 2 ( ) ( ) = ± 2 2 2 2 0 0 t t = + t t 0 0 2 ( ive sign being absurd) = + ( ) 1 2 0 t =2141 0 . t From the begining of the motion the point mass will return to the initial position after time 3 141 0 . t . 12. 5 2 1015 2 2 = +  u ( ) Þ u 2 325 = (a) For H 0 2 10 2 2 = +  u H ( ) i.e., 20 325 H = or H =1625 . m (b) For t 0 325 10 = +  ( )t \ t = 325 10 =18 . s 13. (a) 15 2 60 2 2 = + ´ u a …(i) and 15 6 = + ´ u a …(ii) Substituting the value of 6a from Eq. (ii) in Eq. (i) 225 20 15 2 = +  u u ( ) 22  Mechanics1 T O W E R H A 5m 25 m u = 0 B Time interval t second for A (t–1) second for B A B 1 s H 15m v = 0 5 m/s u x 60 m 6.0 s At rest a u a 15 m/s Page 4 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s =  64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \  = +  64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t   = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = +  ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = 60m 6.1s = 9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration =  + Final velocity initial velocity s ( ) 4 8 =  0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under vt graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20  Mechanics1 h +20 m/s – 60 m v time 12s 8s 4s v max q (b)As the particle did not return back distance travelled in 12 s = Displacement at 12 s \ Average speed = 8 m/s. 6. (a) Radius ( ) R of circle = 21 22 m \ Circumference of circle = 2pR = ´ ´ 2 22 7 21 22 m =6 m Speed ( ) v of particle = 1 m/s \ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved = ´ = = ° 2 6 2 2 3 120 p p Magnitude of Average velocity = = Magnitude of displacement Time s ( ) 2 = = ° AB R 2 2 60 2 sin =R 3 2 = 21 22 3 2 = 21 3 44 m/s (b) Magnitude of average acceleration =  ½ ½ ½ ½ ½ ½ ½ ½ ® ® v v B A 2s = ° 2 120 2 2 vsin [Q     v v A B v ® ® = = = 1 m/s) = ° vsin 60 = 3 2 m/s 2 7. Position vector at t = 0 s r i j 1 1 2 ® = + ( ) ^ ^ m Position vector at t = 4 s r i j 2 6 4 ® = + ( ) ^ ^ m (a) Displacement from t = 0 s to t = 4 s =  ® ® ( ) r r 2 1 = +  + ( ) ( ) ^ ^ ^ ^ 6 4 1 2 i j i j = + ( ) ^ ^ 5 2 i j m \ Average velocity = + ( ) ^ ^ 5 2 4 i j m s = + ( . . ) ^ ^ 125 05 i j m/s (b) Average acceleration =  Final velocity Initial velocity s 4 = +  + ( ) ( ) ^ ^ ^ ^ 2 10 4 6 4 i j i j =  + 2 4 4 i j ^ ^ =  + ( . ) ^ ^ 05i j m/s 2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement between A and B is 10 m 1 2 1 2 1 10 2 2 gt gt   = ( ) or t t 2 2 1 2   = ( ) or t t t 2 2 2 1 2   + = ( ) or 2 3 t = t =1.5 s 9. The two bodies will meet if Displacement of first after attaining = highest point Displacement of second before attaining highest point Motion in One Dimension  21 120° O B A(t = 0 s) (t = 2s) v B ® v A ® v t gt v t t g t t 0 2 0 0 0 2 1 2 1 2  =    ( ) ( ) or 0 1 2 2 0 0 0 2 0 =    v t g t t t ( ) or gt t gt v t 0 0 2 0 0 1 2 = + or t gt v g = + 1 2 0 0 = + t v g 0 0 2 10. 5 1 2 2 = gt Þ t =1s For A H t gt = + 0 1 2 2 . or H gt = 1 2 2 …(i) For B H g t  =  25 1 2 1 2 ( ) …(ii) or 1 2 1 2 1 25 2 2 gt gt   = ( ) [Substituting value of H from Eq. (i)] 1 2 1 25 2 2 g t t [ ( ) ]   = t t t 2 2 2 1 5   + = ( ) 2 1 5 t  = Þ t = 3 s Substituting t = 3 s in Eq. (i) H = ´ ´ = 1 2 10 3 45 2 m 11. s at = 1 2 0 2 Forward motion v at = 0 Backward mo tion  = +  s at t a t ( ) ( ) 0 2 1 2 \  =  1 2 1 2 0 2 0 2 at at t at ( ) or  =  t t t t 0 2 0 2 2 or t t t t 2 0 0 2 2 0   = t t t t = ±    2 2 41 2 0 0 2 0 2 ( ) ( ) = ± 2 2 2 2 0 0 t t = + t t 0 0 2 ( ive sign being absurd) = + ( ) 1 2 0 t =2141 0 . t From the begining of the motion the point mass will return to the initial position after time 3 141 0 . t . 12. 5 2 1015 2 2 = +  u ( ) Þ u 2 325 = (a) For H 0 2 10 2 2 = +  u H ( ) i.e., 20 325 H = or H =1625 . m (b) For t 0 325 10 = +  ( )t \ t = 325 10 =18 . s 13. (a) 15 2 60 2 2 = + ´ u a …(i) and 15 6 = + ´ u a …(ii) Substituting the value of 6a from Eq. (ii) in Eq. (i) 225 20 15 2 = +  u u ( ) 22  Mechanics1 T O W E R H A 5m 25 m u = 0 B Time interval t second for A (t–1) second for B A B 1 s H 15m v = 0 5 m/s u x 60 m 6.0 s At rest a u a 15 m/s i.e., u u 2 20 75 0  + = ( )( ) u u   = 15 5 0 \ u = 5 m/s (15 m/s being not possible) (b) Using Eq. (ii) a = 5 3 m / s 2 (c) u ax 2 2 0 2 = + i.e., x u a = = ´ 2 2 2 5 2 5 3 ( ) =75 . m (d) s at = 1 2 2 = ´ ´ 1 2 5 3 2 t = 5 6 2 t …(iii) t( ) s 0 1 3 6 9 12 v (m/s) 0 5/6 7.5 30 67.5 120 Differentiating Eq. (iii) w.r.t. time t v t = 5 3 t( ) s 0 3 6 9 12 v (m/s) 0 5 10 15 20 14. Jour ney A to P v xt max = + 0 1 …(i) and v xs max 2 1 2 = …(ii) Þ t v x 1 = max Jour ney P to B 0 2 = +  v y t max ( ) …(iii) and v ys max 2 2 2 = …(iv) Þ t v y 2 = max \ v x v y t t max max + = + = 1 2 4 or v x y max 1 1 4 + é ë ê ù û ú = …(v) From Eq. (ii) and Eq. (iv) s s v x v y 1 2 2 2 2 2 + = + max max or 4 2 1 1 2 = + é ë ê ù û ú v x y max …(vi) Dividing Eq. (vi) by Eq. (v) v max = 2 Substituting the value of v max in Eq. (v) 1 1 2 x y + = (Proved) 15. Let acceleration of the particle be a using v u at = + 0 6 = + u a \ a u =  6 (a) At t = 10 s, s =  2 m  = ´ + ´ 2 10 1 2 10 2 u a or  =  + 2 6 10 50 ( ) a a or  =  10 2 a or a = 0.2 m/s 2 Motion in One Dimension  23 20 15 10 5 3 6 9 12 v (m/s) t (s) Starts a = + x t 1 t 2 a = –y Stops 4 km B s 1 s 2 4 min B A 120 90 60 30 15 3 6 9 12 t (s) s (m) 2m 4 0 t = 10 s t = 6 s t = 0 s v = 0 +ive xaxis Page 5 AIEEE Corner Subjective Questions (Level 1) 1. (a) D v t v t = + 1 1 2 2 = ´ æ è ç ö ø ÷ + ´ æ è ç ö ø ÷ 60 1 80 1 2 km h h km h h =100 km (b) Average speed = = 100km 1.5h 66.67 km/h. 2. (a) Displacement in first two seconds = + ( )( ) ( ) 4 2 1 2 6 2 2 =20 m \ Average velocity = = 20 2 10 m s m/s (b) Displacement in first four seconds = + ( )( ) ( ) 4 4 1 2 6 4 2 =64 m \ Displacement in the time interval t =2 s to t = 4 s =  64 20 = 44 m \ Average velocity = = 44 2 22 m s m/s. 3. Let the particle takes t time to reach ground. \  = +  64 20 1 2 10 2 t t ( ) i.e., 5 20 64 0 2 t t   = t = 6.1 s If the particle goes h meter above tower before coming down 0 20 2 10 2 = +  ( ) ( ) h Þ h = 20 m (a) Average speed = Total distance moved Time taken = + + ( ) 20 20 60 6.1 =16.4 m/s (b) Average velocity = Net displacement Time taken = 60m 6.1s = 9.8 m/s =9.8 m/s (downwards) 4. Average velocity = S S vt t 2.5v vt vt vT t t T = + + + + 0 0 0 0 2 3 \ 5 3 3 0 0 t T t T + = + 2.5 or T t = 4 0 5. (a) Average acceleration =  + Final velocity initial velocity s ( ) 4 8 =  0 0 12 =0 m/s 2 (c) a v = = tan max q 4 \ 4 4 = v max (Q a = 4 m/s 2 ) i.e., v max =16 m/s Displacement of particle in 12 seconds = Area under vt graph = ´ 12 2 v max = ´ 12 16 2 =96 m Average velocity = Displacement m) at Time ( ( ) 96 12 12 s s = +8 m/s 20  Mechanics1 h +20 m/s – 60 m v time 12s 8s 4s v max q (b)As the particle did not return back distance travelled in 12 s = Displacement at 12 s \ Average speed = 8 m/s. 6. (a) Radius ( ) R of circle = 21 22 m \ Circumference of circle = 2pR = ´ ´ 2 22 7 21 22 m =6 m Speed ( ) v of particle = 1 m/s \ Distance moved by particle in 2 s = 2 m Thus, angle through which the particle moved = ´ = = ° 2 6 2 2 3 120 p p Magnitude of Average velocity = = Magnitude of displacement Time s ( ) 2 = = ° AB R 2 2 60 2 sin =R 3 2 = 21 22 3 2 = 21 3 44 m/s (b) Magnitude of average acceleration =  ½ ½ ½ ½ ½ ½ ½ ½ ® ® v v B A 2s = ° 2 120 2 2 vsin [Q     v v A B v ® ® = = = 1 m/s) = ° vsin 60 = 3 2 m/s 2 7. Position vector at t = 0 s r i j 1 1 2 ® = + ( ) ^ ^ m Position vector at t = 4 s r i j 2 6 4 ® = + ( ) ^ ^ m (a) Displacement from t = 0 s to t = 4 s =  ® ® ( ) r r 2 1 = +  + ( ) ( ) ^ ^ ^ ^ 6 4 1 2 i j i j = + ( ) ^ ^ 5 2 i j m \ Average velocity = + ( ) ^ ^ 5 2 4 i j m s = + ( . . ) ^ ^ 125 05 i j m/s (b) Average acceleration =  Final velocity Initial velocity s 4 = +  + ( ) ( ) ^ ^ ^ ^ 2 10 4 6 4 i j i j =  + 2 4 4 i j ^ ^ =  + ( . ) ^ ^ 05i j m/s 2 (c) We cannot find the average speed as the actual path followed by the particle is not known. Uniform acceleration (a) One dimensional motion 8. If at time t the vertical displacement between A and B is 10 m 1 2 1 2 1 10 2 2 gt gt   = ( ) or t t 2 2 1 2   = ( ) or t t t 2 2 2 1 2   + = ( ) or 2 3 t = t =1.5 s 9. The two bodies will meet if Displacement of first after attaining = highest point Displacement of second before attaining highest point Motion in One Dimension  21 120° O B A(t = 0 s) (t = 2s) v B ® v A ® v t gt v t t g t t 0 2 0 0 0 2 1 2 1 2  =    ( ) ( ) or 0 1 2 2 0 0 0 2 0 =    v t g t t t ( ) or gt t gt v t 0 0 2 0 0 1 2 = + or t gt v g = + 1 2 0 0 = + t v g 0 0 2 10. 5 1 2 2 = gt Þ t =1s For A H t gt = + 0 1 2 2 . or H gt = 1 2 2 …(i) For B H g t  =  25 1 2 1 2 ( ) …(ii) or 1 2 1 2 1 25 2 2 gt gt   = ( ) [Substituting value of H from Eq. (i)] 1 2 1 25 2 2 g t t [ ( ) ]   = t t t 2 2 2 1 5   + = ( ) 2 1 5 t  = Þ t = 3 s Substituting t = 3 s in Eq. (i) H = ´ ´ = 1 2 10 3 45 2 m 11. s at = 1 2 0 2 Forward motion v at = 0 Backward mo tion  = +  s at t a t ( ) ( ) 0 2 1 2 \  =  1 2 1 2 0 2 0 2 at at t at ( ) or  =  t t t t 0 2 0 2 2 or t t t t 2 0 0 2 2 0   = t t t t = ±    2 2 41 2 0 0 2 0 2 ( ) ( ) = ± 2 2 2 2 0 0 t t = + t t 0 0 2 ( ive sign being absurd) = + ( ) 1 2 0 t =2141 0 . t From the begining of the motion the point mass will return to the initial position after time 3 141 0 . t . 12. 5 2 1015 2 2 = +  u ( ) Þ u 2 325 = (a) For H 0 2 10 2 2 = +  u H ( ) i.e., 20 325 H = or H =1625 . m (b) For t 0 325 10 = +  ( )t \ t = 325 10 =18 . s 13. (a) 15 2 60 2 2 = + ´ u a …(i) and 15 6 = + ´ u a …(ii) Substituting the value of 6a from Eq. (ii) in Eq. (i) 225 20 15 2 = +  u u ( ) 22  Mechanics1 T O W E R H A 5m 25 m u = 0 B Time interval t second for A (t–1) second for B A B 1 s H 15m v = 0 5 m/s u x 60 m 6.0 s At rest a u a 15 m/s i.e., u u 2 20 75 0  + = ( )( ) u u   = 15 5 0 \ u = 5 m/s (15 m/s being not possible) (b) Using Eq. (ii) a = 5 3 m / s 2 (c) u ax 2 2 0 2 = + i.e., x u a = = ´ 2 2 2 5 2 5 3 ( ) =75 . m (d) s at = 1 2 2 = ´ ´ 1 2 5 3 2 t = 5 6 2 t …(iii) t( ) s 0 1 3 6 9 12 v (m/s) 0 5/6 7.5 30 67.5 120 Differentiating Eq. (iii) w.r.t. time t v t = 5 3 t( ) s 0 3 6 9 12 v (m/s) 0 5 10 15 20 14. Jour ney A to P v xt max = + 0 1 …(i) and v xs max 2 1 2 = …(ii) Þ t v x 1 = max Jour ney P to B 0 2 = +  v y t max ( ) …(iii) and v ys max 2 2 2 = …(iv) Þ t v y 2 = max \ v x v y t t max max + = + = 1 2 4 or v x y max 1 1 4 + é ë ê ù û ú = …(v) From Eq. (ii) and Eq. (iv) s s v x v y 1 2 2 2 2 2 + = + max max or 4 2 1 1 2 = + é ë ê ù û ú v x y max …(vi) Dividing Eq. (vi) by Eq. (v) v max = 2 Substituting the value of v max in Eq. (v) 1 1 2 x y + = (Proved) 15. Let acceleration of the particle be a using v u at = + 0 6 = + u a \ a u =  6 (a) At t = 10 s, s =  2 m  = ´ + ´ 2 10 1 2 10 2 u a or  =  + 2 6 10 50 ( ) a a or  =  10 2 a or a = 0.2 m/s 2 Motion in One Dimension  23 20 15 10 5 3 6 9 12 v (m/s) t (s) Starts a = + x t 1 t 2 a = –y Stops 4 km B s 1 s 2 4 min B A 120 90 60 30 15 3 6 9 12 t (s) s (m) 2m 4 0 t = 10 s t = 6 s t = 0 s v = 0 +ive xaxis (b) v t u a ( ) at s = = + 10 10 =  + 6 10 a a = 4a =08 . m/s (c) Two or three dimensional motion 16. a F ® ® = = m 10 2 N north kg =5 m/s 2 , north =5j ^ m/s 2 u ® =10 m/s, east =10i ^ m/s using v u a ® ® ® = + t v i j ® = + ´ 10 5 2 ^ ^ ( ) = + 10 10 i j ^ ^   v ® =10 2 m/s v ® =10 2, northeast using s u a ® ® ® = + t t 1 2 2 = ´ + ( ) ( ) ^ ^ 10 2 1 2 5 2 2 i j = + ( ) ^ ^ 20 10 i j m   s ® = + 20 10 2 2 =10 5 m cotq = = 20 10 2 q =  cot 1 2 s ® =10 5 m at cot ( ) 1 2 from east to north. 17. s i j ® = + 0 2 4 ( ) ^ ^ m a i ® = 1 2 ^ m/s 2 (t = 0 s to t = 2 s) t 1 2 = s u 0 ® ® = m/s a j ® =  2 4 ^ m/s 2 (t = 2 s to t = 4 s) t 2 2 = s (a) Velocity v u a ® ® ® = + 1 1 1 t = + ® 0 i ( ) ^ 2 2 = 4i ^ v v a a ® ® ® ® = + + 2 1 1 2 2 ( )t = +  4 2 4 2 i i j ^ ^ ^ ( ) or v i j ® =  2 8 8 ( ) ^ ^ m/s (b) Coordinate of particle s s u a ® ® ® ® = + + 1 0 1 1 1 2 1 2 t t = + + + ( ) ( )( ) ( ) ^ ^ ^ 2 4 0 2 1 2 2 2 2 i j i = + + 2 4 4 i j i ^ ^ ^ = + 6 4 i j ^ ^ s s v ® ® ® = + + + 2 1 1 2 1 2 2 2 1 2 t a a t ( ) = + + +  6 4 4 2 1 2 2 4 2 2 i j i i j ^ ^ ^ ^ ^ ( ) ( ) = + + +  6 4 8 4 8 i j i i j ^ ^ ^ ^ ^ =  18 4 i j ^ ^ Coordinate of the particle [ , ] 18 4 m m  18. u i j ® =  ( ) ^ ^ 2 4 m/s, s 0 ® ® = 0 m a i j ® = + ( ) ^ ^ 4 m/s 2 (a) Velocity v u a ® ® ® = + t =  + + ( ) ( ) ^ ^ ^ ^ 2 4 4 2 i j i j =  ( ) ^ ^ 10 2 i j m/s (b) Coordinates of the particle s s u a ® ® ® ® = + + 0 1 2 2 t t = +  + + ® 0 i j i j ( ) ( $ ) ^ ^ ^ 2 4 1 2 4 2 2 24  Mechanics1 O North East q s ® j ^ i ^Read More
1. What is motion in one dimension? 
2. How is motion in one dimension different from motion in two or three dimensions? 
3. What are the key equations used to describe motion in one dimension? 
4. How can we calculate the velocity and acceleration of an object in onedimensional motion? 
5. What are some examples of motion in one dimension? 

Explore Courses for NEET exam
